walters
  • walters
Question based on (Groups)
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

walters
  • walters
jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel \[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\] Determine whether or not G is an abelian group under the composition of mappings
walters
  • walters
oops ignore jjj
KingGeorge
  • KingGeorge
Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

walters
  • walters
am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group
KingGeorge
  • KingGeorge
It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well. Have you shown any of the axioms for proving it's a group yet?
walters
  • walters
i am failing to show the axioms
KingGeorge
  • KingGeorge
First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.
walters
  • walters
why do u chose to use |dw:1360875075213:dw|
KingGeorge
  • KingGeorge
"\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.
walters
  • walters
can u also use * if u wnt
KingGeorge
  • KingGeorge
For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{-1}\) and \(d=a^{-1}(-b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{-1}ax+ba^{-1}-a^{-1}b=x,\]so we do indeed have inverses.
KingGeorge
  • KingGeorge
If you make it clear your group function is composition of functions, you can use * as well.
KingGeorge
  • KingGeorge
As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.
walters
  • walters
please do
KingGeorge
  • KingGeorge
let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.
walters
  • walters
k it is a group but not an abelian group
KingGeorge
  • KingGeorge
Yup.
walters
  • walters
thnx
KingGeorge
  • KingGeorge
You're welcome.
walters
  • walters
can we verify identity
KingGeorge
  • KingGeorge
\[\tau_{10}(x)=1x+0=x\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.